Understanding the Isosceles Trapezium Vertices

We are given an isosceles trapezium named \(ABCD\), where the side \(AB\) is parallel to the side \(DC\). The coordinates for three vertices are provided: \(A(2, 3)\), \(B(4, 3)\), and \(C(5, 1)\). Our first step is to find the coordinates of the fourth vertex, \(D\).

Key properties we'll use:

• \(AB\) is parallel to \(DC\).
• The trapezium is isosceles, meaning the non-parallel sides \(AD\) and \(BC\) are equal in length (\(AD = BC\)).

Analyzing the coordinates:

• Vertices \(A(2, 3)\) and \(B(4, 3)\) have the same y-coordinate (3). This means the side \(AB\) is a horizontal line.
• Since \(AB\) is parallel to \(DC\), the side \(DC\) must also be a horizontal line. Therefore, the y-coordinate of vertex \(D\) must be the same as the y-coordinate of vertex \(C\), which is 1. Let the coordinates of \(D\) be \((x, 1)\).

Now, we use the property that \(AD = BC\).

1. Calculate the square of the distance between \(B\) and \(C\) (\(BC^2\)):

   Using the distance formula \(d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\): \(BC^2 = (5 - 4)^2 + (1 - 3)^2\) \(BC^2 = (1)^2 + (-2)^2\) \(BC^2 = 1 + 4\) \(BC^2 = 5\)

2. Calculate the square of the distance between \(A\) and \(D\) (\(AD^2\)), where \(D = (x, 1)\):

   \(AD^2 = (x - 2)^2 + (1 - 3)^2\) \(AD^2 = (x - 2)^2 + (-2)^2\) \(AD^2 = (x - 2)^2 + 4\)

3. Set \(AD^2 = BC^2\):

   \((x - 2)^2 + 4 = 5\) \((x - 2)^2 = 5 - 4\) \((x - 2)^2 = 1\)

4. Solve for \(x\):

   Taking the square root of both sides: \(x - 2 = ±1\).

   • Case 1: \(x - 2 = 1\) => \(x = 3\). This would make \(D = (3, 1)\). In this case, the length of \(DC\) would be \(|5-3|=2\), same as \(AB\). This forms a parallelogram, a special case of an isosceles trapezium.
   • Case 2: \(x - 2 = -1\) => \(x = 1\). This makes \(D = (1, 1)\). Here, the length of \(DC\) is \(|5-1|=4\), different from \(AB\). This fits the standard definition of an isosceles trapezium where non-parallel sides are equal.

   Given the context and typical geometry problems, we choose the configuration where \(AB\) and \(DC\) are the parallel bases of different lengths. Thus, we select \(x = 1\).

The coordinates of vertex \(D\) are \((1, 1)\).

Finding the Diagonal Intersection Point

The diagonals of the trapezium are the line segments \(AC\) and \(BD\). To find their intersection point, we first determine the equations of the lines containing these diagonals.

Equation for Diagonal AC

The line passes through \(A(2, 3)\) and \(C(5, 1)\).

1. Calculate the slope (\(m_{AC}\)):

   \(m_{AC} = \frac{1 - 3}{5 - 2} = \frac{-2}{3}\)

2. Use the point-slope form (\(y - y_1 = m(x - x_1)\)) with point \(A(2, 3)\):

   \(y - 3 = -\frac{2}{3}(x - 2)\)

3. Convert to the standard form (\(ax + by = c\)):

   \(3(y - 3) = -2(x - 2)\)

   \(3y - 9 = -2x + 4\)

   \(2x + 3y = 4 + 9\)

   Equation 1: \(2x + 3y = 13\)

Equation for Diagonal BD

The line passes through \(B(4, 3)\) and \(D(1, 1)\).

1. Calculate the slope (\(m_{BD}\)):

   \(m_{BD} = \frac{1 - 3}{1 - 4} = \frac{-2}{-3} = \frac{2}{3}\)

2. Use the point-slope form with point \(B(4, 3)\):

   \(y - 3 = \frac{2}{3}(x - 4)\)

3. Convert to the standard form:

   \(3(y - 3) = 2(x - 4)\)

   \(3y - 9 = 2x - 8\)

   \(2x - 3y = -9 + 8\)

   Equation 2: \(2x - 3y = -1\)

Solving the System of Equations

We need to find the point \((x, y)\) that satisfies both equations:

1. \(2x + 3y = 13\)
2. \(2x - 3y = -1\)

We can use the elimination method. Add Equation 1 and Equation 2:

\((2x + 3y) + (2x - 3y) = 13 + (-1)\)

\(4x = 12\)

\(x = 12 / 4\)

\(x = 3\)

Now, substitute the value of \(x = 3\) into Equation 1:

\(2(3) + 3y = 13\)

\(6 + 3y = 13\)

\(3y = 13 - 6\)

\(3y = 7\)

\(y = 7 / 3\)

The point of intersection of the diagonals \(AC\) and \(BD\) is \((3, 7/3)\).

Conclusion on Intersection Point

By determining the coordinates of the fourth vertex \(D\) using the properties of an isosceles trapezium and then finding the equations of the diagonals \(AC\) and \(BD\), we solved the system of linear equations. The calculation shows that the diagonals intersect at the point \((3, 7/3)\).